Chapter 21: RLC · PDF fileChapter 21: RLC Circuits. PHY2054: Chapter 21 2 ... ÎThe figure shows the current and emf of a series RLC circuit. To increase the rate at which power is

PHY2054: Chapter 21 1

Chapter 21: RLC Circuits


Voltage and Current in RLC CircuitsAC emf source: “driving frequency” f

If circuit contains only R + emf source, current is simple

If L and/or C present, current is not in phase with emf

Z, φ shown later

( )sin mm mi I t I

Zε

ω φ= − =

sinm tε ε ω= 2 fω π=

( ) ( )sin current amplitudemm mi I t I

R Rεε ω= = =


AC Source and Resistor OnlyDriving voltage is

Relation of current and voltage

Current is in phase with voltage (φ = 0)

i

ε R~sin m

m mi I t IRε

ω= =

sinm tε ε ω=

/i Rε=


AC Source and Capacitor OnlyVoltage is

Differentiate to find current

Rewrite using phase (check this!)


“Capacitive reactance”:Current “leads” voltage by 90°

sinmq C tε ω= i

ε C~/ cosCi dq dt CV tω ω= =

sinC mqv tC

ε ω= =

( )sin 90Ci CV tω ω= + °

( )sin 90 mm m

Ci I t I

Xεω= + ° =

1/CX Cω=

( )1/CX Cω=


AC Source and Inductor OnlyVoltage is

Integrate di/dt to find current:

Rewrite using phase (check this!)


“Inductive reactance”:Current “lags” voltage by 90°

( )/ / sinmdi dt L tε ω= i

ε L~( )/ cosmi L tε ω ω= −

/ sinL mv Ldi dt tε ω= =

( ) ( )/ sin 90mi L tε ω ω= − °

( )sin 90 mm m

Li I t I

Xεω= − ° =

LX Lω=

( )LX Lω=


General Solution for RLC CircuitWe assume steady state solution of form

Im is current amplitudeφ is phase by which current “lags” the driving EMFMust determine Im and φ

Plug in solution: differentiate & integrate sin(ωt-φ)

( )sinmi I tω φ= −

( ) ( ) ( )cos sin cos sinmm m m

II L t I R t t tC

ω ω φ ω φ ω φ ε ωω

− + − − − =

sinmdi qL Ri tdt C

ε ω+ + =

( )sinmi I tω φ= −

( )cosmdi I tdt

ω ω φ= −

( )cosmIq tω φω

= − −

Substitute


General Solution for RLC Circuit (2)

Expand sin & cos expressions

Collect sinωt & cosωt terms separately

These equations can be solved for Im and φ (next slide)

( )( )

1/ cos sin 0

1/ sin cosm m m

L C R

I L C I R

ω ω φ φ

ω ω φ φ ε

− − =

− + =

( )( )

sin sin cos cos sin

cos cos cos sin sin

t t t

t t t

ω φ ω φ ω φ

ω φ ω φ ω φ

− = −

− = +High school trig!

cosωt terms

sinωt terms

( ) ( ) ( )cos sin cos sinmm m m

II L t I R t t tC

ω ω φ ω φ ω φ ε ωω

− + − − − =


Solve for φ and Im

R, XL, XC and Z have dimensions of resistance

This is where φ, XL, XC and Z come from!

General Solution for RLC Circuit (3)

1/tan L CX XL CR R

ω ωφ −−= ≡ m

mIZε

=

( )22L CZ R X X= + −

LX Lω=

1/CX Cω=

Inductive “reactance”

Capacitive “reactance”

Total “impedance”


AC Source and RLC Circuits

tan

mm

L C

IZ

X XR

ε

φ

=

−= Phase angle

Maximum current

φ= angle that current “lags” applied voltage

( )21/

L

C

X L fX C

ω ω πω

= =

=

Inductive reactance

Capacitive reactance

( )22L CZ R X X= + − Total impedance


What is Reactance?Think of it as a frequency-dependent resistance

Shrinks with increasing ω1

CXCω

=

LX Lω=

( " " )RX R=

Grows with increasing ω

Independent of ω


Pictorial Understanding of Reactance

tan L CX XR

φ −=

( )22L CZ R X X= + −

cos RZ

φ =


Summary of Circuit Elements, Impedance, Phase Angles

( )22L CZ R X X= + − tan L CX X

Rφ −=


QuizThree identical EMF sources are hooked to a single circuit element, a resistor, a capacitor, or an inductor. The current amplitude is then measured as a function of frequency. Which one of the following curves corresponds to an inductive circuit?

(1) a(2) b(3) c(4) Can’t tell without more info

f

Imax

a

c

b

( )max max

2/

L

L

X L fI X

ω ω πε

= =

=For inductor, higher frequency gives higherreactance, therefore lower current


RLC Example 1Below are shown the driving emf and current vs time of an RLC circuit. We can conclude the following

Current “leads” the driving emf (φ<0)Circuit is capacitive (XC > XL)

εI

t


RLC Example 2R = 200Ω, C = 15μF, L = 230mH, εmax = 36v, f = 60 Hz

2 60 0.23 86.7LX π= × × = Ω

( )61/ 2 60 15 10 177CX π −= × × × = Ω

( )22200 86.7 177 219Z = + − = Ω

max max / 36 / 219 0.164AI Zε= = =

XC > XLCapacitive circuit

1 86.7 177tan 24.3200

φ − −⎛ ⎞= = − °⎜ ⎟⎝ ⎠

Current leads emf(as expected)

( )0.164sin 24.3i tω= + °


ResonanceConsider impedance vs frequency

Z is minimum whenThis is resonance!

At resonanceImpedance = Z is minimumCurrent amplitude = Im is maximum

( ) ( )2 22 2 1/L CZ R X X R L Cω ω= + − = + −

1/L Cω ω= 0 1/ LCω ω= =


Imax vs Frequency and ResonanceCircuit parameters: C = 2.5μF, L = 4mH, εmax = 10v

f0 = 1 / 2π(LC)1/2 = 1590 HzPlot Imax vs f

R = 5Ω

R = 10Ω

R = 20ΩImax

Resonance0f f=

f / f0

( )22max 10 / 1/I R L Cω ω= + −


Power in AC CircuitsInstantaneous power emitted by circuit: P = i2R

More useful to calculate power averaged over a cycleUse <…> to indicate average over a cycle

Define RMS quantities to avoid ½ factors in AC circuits

House currentVrms = 110V ⇒ Vpeak = 156V

( )2 2sinm dP I R tω φ= −

( )2 2 212sinm d mP I R t I Rω φ= − =

rms 2mII = rms 2

mεε = 2ave rmsP I R=

Instantaneous power oscillates


Power in AC CircuitsPower formula

Rewrite using

cosφ is the “power factor”To maximize power delivered to circuit ⇒ make φ close to zeroMax power delivered to load happens at resonanceE.g., too much inductive reactance (XL) can be cancelled by increasing XC (e.g., circuits with large motors)

2ave rmsP I R=

rmsave rms rms rms cos

ZP I R Iε ε φ= =

ave rms rms cosP Iε φ= cos RZ

φ =

rmsrmsI

Zε

=

rms max / 2I I=

R

L CX X−Z

φ


Power Example 1R = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz

( )22200 80 150 211.9Z = + − = Ω

1 80 150tan 19.3200

φ − −⎛ ⎞= = − °⎜ ⎟⎝ ⎠

cos 0.944φ =

ave rms rms cos 120 0.566 0.944 64.1WP Iε φ= = × × =

rms rms / 120 / 211.9 0.566AI Zε= = =

2 2ave rms 0.566 200 64.1WP I R= = × =

Current leads emfCapacitive circuit

Same


Power Example 1 (cont)R = 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, f = 60 Hz

How much capacitance must be added to maximize the power in the circuit (and thus bring it into resonance)?

Want XC = XL to minimize Z, so must decrease XC

So we must add 15.5μF capacitance to maximize power

150 1/ 2 17.7μFCX fC Cπ= Ω = =

new new80 33.2μFC LX X C= = Ω =


Power vs Frequency and ResonanceCircuit parameters: C = 2.5μF, L = 4mH, εmax = 10v

f0 = 1 / 2π(LC)1/2 = 1590 HzPlot Pave vs f for different R values

R = 5Ω

R = 10Ω

R = 20Ω0f f=

Pave

R = 2Ω

Resonance

f / f0


Resonance Tuner is Based on ResonanceVary C to set resonance frequency to 103.7 (ugh!)

Circuit response Q = 500 Tune for f = 103.7 MHzOther radio stations.

RLC response is less


QuizA generator produces current at a frequency of 60 Hz with peak voltage and current amplitudes of 100V and 10A, respectively. What is the average power produced if they are in phase?

(1) 1000 W(2) 707 W(3) 1414 W(4) 500 W(5) 250 W

1ave peak peak rms rms2P I Iε ε= =


QuizThe figure shows the current and emf of a series RLC circuit. To increase the rate at which power is delivered to the resistive load, which option should be taken?

(1) Increase R(2) Decrease L(3) Increase L(4) Increase C

Current lags applied emf (φ > 0), thus circuit is inductive. Either(1) Reduce XL by decreasing L or(2) Cancel XL by increasing XC (decrease C).

tan L CX XR

φ −=


Example: LR CircuitVariable frequency EMF source with εm=6V connected to a resistor and inductor. R=80Ω and L=40mH.

At what frequency f does VR = VL?

At that frequency, what is phase angle φ?

What is the current amplitude and RMS value?

2000 2000 / 2 318HzLX L R fω ω π= = ⇒ = = =

tan / 1 45LX Rφ φ= = ⇒ = °

2 2max max / 80 80 6 /113 0.053AI ε= + = =

rms max / 2 0.037AI I= =

( )0.053sin 45i tω= − °


TransformersPurpose: change alternating (AC) voltage to a bigger (or smaller) value

p pBV Nt

ΔΦ=

Δ

Bs sV N

tΔΦ

=Δ

Input AC voltagein the “primary”turns produces a flux

ss p

p

NV VN

=

Changing flux in“secondary” turnsinduces an emf


TransformersNothing comes for free, however!

Increase in voltage comes at the cost of current.Output power cannot exceed input power!power in = power out(Losses usually account for 10-20%)

p p s si V i V=

p ps

p s s

V Nii V N

= =


Transformers: Sample Problem A transformer has 330 primary turns and 1240 secondary turns. The input voltage is 120 V and the output current is 15.0 A. What is the output voltage and input current?

1240120 451V330

ss p

p

NV VN

⎛ ⎞= = =⎜ ⎟⎝ ⎠

“Step-up”transformer

p p s si V i V=45115 56.4A120

sp s

p

Vi iV

⎛ ⎞= = =⎜ ⎟⎝ ⎠


This is how first experiment by Faraday was done

He only got a deflection of the galvanometer when the switch is opened or closed

Steady current does not make induced emf.

Transformers


Microphone

Tape recorder

Applications


ConcepTest: Power lines At large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit (high V, low i) or (high i, low V)?

(1) high V, low i(2) low V, high i(3) makes no difference

Power loss is i2R


Electric Power Transmission

i2R: 20x smaller current ⇒ 400x smaller power loss

Documents

Chapter 21: RLC · PDF fileChapter 21: RLC Circuits. PHY2054: Chapter 21 2 ... ÎThe figure shows the current and emf of a series RLC circuit. To increase the rate at which power is